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Homework Help: Having trouble with basic problem

  1. Aug 5, 2012 #1
    I'm currently reading Lang's "First course in calculus" and can't seem to find the solution to a rather basic problem.

    "Water is flowing into a tank in the form of a hemisphere of a radius of 10ft with flat side up at the rate of 4ft³/min. Let h be the depth of the water, r the radius of the surface of the water and V the volume of the water in the tank. Assume that [itex]\frac{dV}{dt}=\pi r^2 \frac{dh}{dt}[/itex]. Find how fast the water level is rising when h=5ft."

    So basically I'm trying to find [itex]\frac{dh}{dt}[/itex] when h=5.

    First of all I'm wondering how he got [itex]\frac{dV}{dt}=\pi r^2 \frac{dh}{dt}[/itex], when I tried to get [itex]\frac{dV}{dt}[/itex] I got, since:
    [tex]V=\frac{2}{3} \pi r^2 h[/tex]

    so by the chain rule:
    [tex]\frac{dV}{dt}=\frac{2}{3} \pi [r^2 \frac{dh}{dt}+2r\frac{dr}{dt}h][/tex]

    What did I do wrong there?

    Also I can't find a way to solve the problem. All I got was:

    Since water flows into the tank at 4ft³/min, [itex]\frac{dV}{dt}=4[/itex]

    Then I could solve for [itex]\frac{dh}{dt}[/itex], so that [itex]4=\pi 10^2 \frac{dh}{dt}[/itex] or [itex]\frac{dh}{dt}=\frac{1}{25 \pi}[/itex].

    Completely confused right now. Am I completely wrong? What am I missing here?
  2. jcsd
  3. Aug 5, 2012 #2
    Your formula for the volume is not correct for every [itex]r[/itex] and [itex]h[/itex]. Think about when the hemisphere is only partially full. [itex]r[/itex] is not the radius of the hemisphere but of the circle that the surface of the water forms. He's giving you [itex]dV/dt[/itex] precisely so you don't have to do all the geometry to figure out what the actual volume function is.

    Also, you plugged in [itex]r=5[/itex] instead of [itex]h=5[/itex].
  4. Aug 5, 2012 #3
    Ah ok, now I get it. So I would have to find the radius of the water surface at a height of 5, which I could find with Phytagorean Theorem and then just plug it in.

    Out of interest though, how did he get the actual volume function?
  5. Aug 5, 2012 #4


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    For a small increase in depth, dh, when the surface area is πr2, what would the increase in volume, dV, be?
  6. Aug 5, 2012 #5
    Yeah, calculating the actual volume isn't too painful (you do some geometry, spherical sector minus a cone is the volume you want), but yeah, what's being done is he's making a simple approximation for the volume change.
  7. Aug 6, 2012 #6
    Well I'm not sure. For a given volume, the surface area is also given, but when you change the depth, the surface area also changes. If the volume is given, you can use this formula [itex]V=\frac{2}{3} \pi r^2 h[/itex], correct?

    Now to find the increase in volume dV when I change the depth by a small amount dh I have to take dV/dh right? Now I'm wondering if I'm not looking for dV/dt, but either for dV/dr or dV/dh, do I actually have a function with two variables r and h? And how do I consider the fact that when I change dh, r also changes?
  8. Aug 6, 2012 #7
    You can't use that formula; that formula uses the actual radius of the hemisphere mungled with the height. It makes no sense.

    What haruspex is saying is that you can approximate the change in volume [itex]dV[/itex] in the following way:

    change in volume = (current area)x(change in height) + (change in area)x(current height)

    This is basically the product rule, right? The first term gives the formula for [itex]dV/dt[/itex] that you were given. The second term was neglected (why, I don't know, but probably just to make your life easier.)
  9. Aug 6, 2012 #8
    Ok thanks, that cleared some things up.

    I have one question left though. If you use the general form of the product rule [itex]\frac{d(fg)}{dx}=f(x)\frac{dg}{dx} + \frac{df}{dx} g(x)[/itex], both f and g depend on x, but with change in volume = (current area)x(change in height) + (change in area)x(current height) the change in the area depends on r and the change in height on h, why can r and h be treated as one variable?
  10. Aug 6, 2012 #9
    They're not being treated as one variable; the changes are with respect to any third variable (the most natural one here being t).
  11. Aug 6, 2012 #10


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    The second term would be of interest if the vessel were expanding sideways. In the present instance, the change in volume is somewhere between (current area)x(change in height) and (new area)x(change in height). For very small dh, these are indistinguishable.
  12. Aug 7, 2012 #11


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    It's important to note that in the differential form, [itex]dV/dh = \pi r^2 [/itex], what the author did is not an approximation, it is exact!
    Last edited: Aug 7, 2012
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